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If a group has a cyclic-factor subnormal series of length $n$, choose one lift of a generator of each nontrivial factor. Induction gives $G_i=⟨G_{i-1},g_i⟩$, so the chosen lifts generate $G$. Thus at most $n$ generators are needed. Applying this to the intersection chain in subgroups of polycyclic groups are polycyclic bounds the number of generators of every subgroup by the same series length.

First establish the two general claims about polycyclic groups. Intersect a subnormal cyclic-factor series $(G_i)$ with $H\le G$. Then $H\cap G_i⊴H\cap G_{i+1}$, and the inclusion induces an injective homomorphismA subgroup of a cyclic group is cyclic, so this is a cyclic-factor subnormal series for $H$, after omitting repetitions. This proves subgroups of polycyclic groups are polycyclic.

To prove the generator bound, choose $g_i\in G_i$ whose coset generates $G_i/G_{i-1}$ whenever that factor is nontrivial. Inductively $G_i=⟨G_{i-1},g_i⟩$, so $G=⟨g_1,\dots ,g_n⟩$. This is the generator bound from a cyclic-factor series: at most $n$ generators suffice.

For the first example, take $G=C_2^m$ with its coordinate flagIt has length exactly $m$ and each factor is $C_2$, so all factors are cyclic.